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Math Help - y'=(x^2+y^2)^(3/2)

  1. #1
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    Oct 2010
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    y'=(x^2+y^2)^(3/2)

    can anybody solve this equation?
    y'=(x^2+y^2)^(3/2)

    thanks
    Last edited by mr fantastic; October 21st 2010 at 04:00 AM. Reason: Re-titled.
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  2. #2
    Senior Member
    Joined
    Mar 2010
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    <br />
x=r(\varphi) \: cos\varphi<br />
    <br />
y=r(\varphi) \: sin\varphi<br />

    <br />
\displaystyle { \frac{dx}{d\varphi}=r' \: cos \varphi \: - \: r \: sin \varphi }<br />

    <br />
\displaystyle {  \frac{dy}{d\varphi}=r' \: sin \varphi \: + \: r \: cos \varphi }<br />

    <br />
 \displaystyle { \frac{ dy}{dx} = \frac{r' \: sin \varphi \: + \: r \: cos \varphi}{r' \: cos \varphi \: - \: r \: sin \varphi}=r^3 }<br />

    <br />
\displaystyle {  r' = \frac{r(r^3 \: sin \varphi + cos \varphi)}{r^3 \: cos \varphi - sin \varphi} }<br />

    Not helps much.
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